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3 years ago

How are instantaneous speed and average speed related for an object with constant speed?

Physics

2 answers:

Gala2k [10]3 years ago

7 0

In the case of an object with constant speed, the instantaneous speed and average speed are equal.

kirza4 [7]3 years ago

4 0

If the speed is constant, then it never changes.
Instantaneous speed is always the same, no matter at what moment you measure it, and average speed during any period of time is also the same number.

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why should society be concerned about continental drift and plate tectonics? What value does this research have for the public i

AleksAgata [21]

Humans should be concerned with continental drift and plate tectonics because these forces have a major impact on the environment and also lead to major natural disasters. Shifts in tectonic plates lead to earthquakes which can also cause things like typhoons or tsunamis. Therefore, begin knowledgeable about the locations of fault lines and areas with the high potential for natural disasters as a result of shifting plates is very important.

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3 years ago

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a cheerleader throws herself into the air with a velocity of 12 m/s at an angle of 75 degrees above the horizontal. what are the

NISA [10]

Answer:

Vx = 3.10 [m/s]

Vy = 11.59 [m/s]

Explanation:

To solve this problem we must decompose the velocity vector by means of the angle on the horizontal.

v = 12 [m/s]

Vx = 12*cos (75) = 3.10 [m/s]

Vy = 12*sin (75) = 11.59 [m/s]

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4 years ago

What is the correct answer?

Anna35 [415]

Answer:3

Explanation:

8 0

3 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago

A sound wave has a frequency of 776 Hz in air and a wavelength of 0.52 m. What is the temperature of the air? Relate the speed o

OLEGan [10]

Answer:

142.7650889

Explanation:

I think the answer above is correct. So to find the velocity of the wave you can do: frequency*wavelength = velocity

in your case, the velocity is equal to 403.5m/s

so we know that the 403.52 is the total velocity of the wave

and the equation: velocity at 0 degrees Celsius * $\sqrt{{T}/{273}}$ where T represents temperature in Kelvins = veloctity

so we set 403.52 = 327* $\sqrt{{T}/{273}}$ and solve for T

T will then equal the degree in Kelvins

so to convert from Kelvins to degrees celsius you subtract 273 from the Kelvins value and you are left with the degrees in Celsius which equals 142.7650889 degrees Celsius

5 0

3 years ago